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Lets say we are given a four components object. To be explicit lets consider that these components are $ x^\mu = \mu $ with $\mu\in{0,1,2,3}$, i.e. $$ x^\mu \sim \left[ \begin{matrix} 0\\ 1\\ 2\\ 3 \end{matrix} \right] .$$ How do we know if these components represent a 4-vector or a spinor? (Forget about the 4-vector typical notation I am using. It's just the notation I choose.) I always read in books that 4-vectors (or tensors in general) are recognized by the way their components transform. Does this also apply to spinors?

Let me expand my question with an "example". Lets consider also some Lorentz transformation parametrized by $\xi_i$ for the "angle" of boosts and $\theta_i$ for the angles of rotations. Suppose we are given the components of this object after the transformation and that they are some array of numbers $y^\mu$. But we are not told how were they calculated or, even more interesting, both $x^\mu$ and $y^\mu$ may were measured. We thus want to relate $x^\mu$ and $y^\mu$.

Suppose that after some 'trial and error' we find that they relate by the linear transformation $$ y^\mu = \left( \exp\left(\frac{i}{2} \omega_{\rho\sigma} \Sigma^{\rho\sigma}\right) \right)^\mu_{\ \ \ \nu} x^\nu $$ (which is nothing more than $y^\mu = \Lambda^\mu_{\ \ \nu} x^\nu$) where $$ \omega_{\rho\sigma} = \left[ \begin{matrix}0 & \xi_{1} & \xi_{2} & \xi_{3}\\ -\xi_{1} & 0 & \theta_{3} & \theta_{2}\\ -\xi_{2} & -\theta_{3} & 0 & \theta_{1}\\ -\xi_{3} & -\theta_{2} & -\theta_{1} & 0 \end{matrix} \right] $$

Is it correct to conclude what follows?

  1. If $\Sigma^{\rho\sigma}$ are given by enter image description here then we conclude that the components represent a 4-vector.

  2. If $\Sigma^{\rho\sigma} = \frac{i}{4} [\gamma^\rho,\gamma^\sigma]$ with $\gamma^\mu$ the Dirac matrices, or more explicitly enter image description here then we conclude that the components represent a spinor.

  3. If the $\Sigma$'s are different from these but satisfy the Lorentz algebra then the components $x^\mu$ represent other type of object different from a 4-vector or a spinor.

Is this correct? If yes, may this be taken as the definition of a spinor (as happens with 4-vectors) independently for they to satisfy or not Dirac equation?

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  • $\begingroup$ You recognize by context (possibly even by dimensional analysis). Dirac Spinor components are "internal states" and have different properties (constraints) than "just" 4-vectors. $\endgroup$ – Alexander Oct 10 '18 at 0:47
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    $\begingroup$ Here's a simple explanation on Wikipedia en.wikipedia.org/wiki/Spin_representation $\endgroup$ – octonion Oct 10 '18 at 1:10
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    $\begingroup$ One way to recognize them is by the way they transform. In general, 4-vectors transform through Lorentz like $\Psi'=M(\Lambda)\Psi$, where $M(\Lambda)$ is the transformation matrix for the Lorentz transformation $\Lambda$. One way to represent this matrix is through rotation generators $M(\Lambda)=exp(\frac{i}{2}\omega^{\mu\nu}J_{\mu\nu})$. On the other hand, and spinor is an object that transform like $\Psi'=exp(\frac{i}{2}\omega^{\mu\nu}S_{\mu\nu})\Psi$, where the generators are given by $S_{\mu\nu}=\frac{i}{4}[\gamma_\mu,\gamma_\nu]$, where $\gamma_\mu$ are the Dirac gamma matrices. $\endgroup$ – Charlie Oct 10 '18 at 1:51
  • $\begingroup$ When represented merely as a column vector, one cannot distinguish them - that's the point of the simple representation! As others have said above, the context should inform you of the nature of the beast. $\endgroup$ – N. Steinle Oct 10 '18 at 2:13
  • $\begingroup$ The Casimirs of your 1 and 2 reps have very differenot eigenvalues... which? $\endgroup$ – Cosmas Zachos Oct 10 '18 at 10:54
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You could say that one can distinguish vectors and spinors by the way they transform. Vectors and spinors belong to different representations of the Lorentz group and therefore have different transformation rules. Your example is indeed correct. So if you know from context how an object transforms, you can guess what it is.

Usually, mathematicians get rather angry when they hear about "defining object X by the way it transforms". How can you transform something you have not defined yet? Still, this is the way many GR textbooks introduce tensors, and from a physics point of view, it is OK. Mathematicians would prefer a bottom-up approach, in which one would start with a group (like the Lorentz group or the rotation group), classify its representations and only then give them silly names like 'spinor' and 'tensor'.

By the way (rather irrelevantly), I think there are some minus signs off in your Lorentz generators for the 4-vector: the boost generators should have two nonzero elements of the same sign.

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  • $\begingroup$ So, if I measure something, then I rotate the experiment and perform the same measurement, analyzing the way the measurements are related I can guess what the nature of that something is, right? For scalars the measurement will yield the same result, and for vectors or spinors (or other objects) the relations between the components will tell me what it is. $\endgroup$ – user171780 Oct 10 '18 at 12:30
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    $\begingroup$ That's right, you got the idea. $\endgroup$ – Stijn B. Oct 10 '18 at 13:16

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